A wheel of radius R = 29.9 cm, mass M = 2.42 kg, and moment of inertia I is moun

Question

A wheel of radius R = 29.9 cm, mass M = 2.42 kg, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure. A light cord wrapped around the wheel supports an object of mass m = 0.691 kg. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord.

Angular Acceleration of a Wheel A wheel of radius R 29.9 cm, mass M 3 2.42 kg, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure. A light cord wrapped around the wheel supports an object of mass m 0.691 kg. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord SOLVE IT Conceptualize: Imagine that the object is a bucket in an old fashioned wishing well. It is tied to a cord that passes around a cylinder equipped with a crank for raising the bucket. After the bucket has been raised, the system is released and the bucket accelerates downward while the cord unwinds off the cylinder Categorize: The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. Analyze: The magnitude of the torque acting on the wheel about its axis of rotation is T 3 TR, where Tis the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by mg the Earth on the wheel and the normal force exerted by the axle An object hangs from a cord on the wheel both pass through the axis of rotation and therefore wrapped around a wheel. produce no torque.) Use the following equation: Ia Solve for a and substitute the net torque: TR (1) a Apply Newton's second law to the motion mg may of the object, taking the downward direction to be positive: Solve for the acceleration a (2) Equations (1) and (2) have three unknowns: a, a, and T. Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the whee rim. Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a Ra.

Explanation / Answer

Using Newton’s 2nd law, you can write equations for both the block and the pulley, solve one of them for T, and plug it in the other one and solve for a. For the block you have:
Using formula,
F = ma = mg - T.....(1)

For the pulley:
Using formula,
= I = RT
MR²a / 2R = RT
Ma / 2 = T........(2)

Plugging (2) into (1),

#we get:

ma = mg - Ma/2
a = mg / (m + M/2)
a = (0.691kg)(9.81m/s²)/[0.691kg + (2.42kg/2)]
a = 11.02m/s²

Then, plugging this into (2),

##we find T to be:

T = (2.42kg)(11.02m/s²) / 2
T = 13.33N

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